Question

Let $\alpha ,\beta $ are the roots of the equation $ax^{2}+bx+c=0$ where $\beta =4\alpha \left(\right.\alpha >0\left.\right)$ . If $3a=2\left(\right.c-b\left.\right)$ and $S=\displaystyle \sum _{r = 0}^{\infty }\beta \left(\left(\alpha \right)^{r}\right)$ , then find the value of $3S$ .

Answer 1

$\because 3=2\left(\frac{c}{a} - \frac{b}{a}\right)$
$\Rightarrow 3=2\left(4 \alpha ^{2} + 5 \alpha \right)\Rightarrow 8\alpha ^{2}+10\alpha -3=0$
$\Rightarrow 8\alpha ^{2}+12\alpha -2\alpha -3=0$
$\Rightarrow \left(\right.2\alpha +3\left.\right)\left(\right.4\alpha -1\left.\right)=0$
$\therefore \alpha =\frac{1}{4}\Rightarrow \beta =4\alpha =1$
$\therefore S=\frac{\beta }{1 - \alpha }=\frac{1}{1 - \frac{1}{4}}=\frac{4}{3}\Rightarrow 3S=4$